If $\frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$, then what would be the value of $a^{b+c} \cdot b^{c+a} \cdot c^{a+b}$? I'm unable to proceed.
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See also Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$ – Martin Sleziak Jan 07 '17 at 18:46
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Let $ \frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k$
$\Rightarrow a=e^{k(b-c)}, b=e^{k(c-a)},c=e^{k(a-b)}$. Then the required value is $ e^{k(b^2-c^2)}. e^{k(c^2-a^2)}. e^{k(a^2-b^2)}=1$
Roby5
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Nitin Uniyal
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Let $$ \frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k\tag{1}$$
From $(1)$, we get
$$\log{a}=k(b-c)\tag{2}$$
$$\log{b}=k(c-a)\tag{3}$$
$$\log{c}=k(a-b)\tag{4}$$
Let $$l=a^{b+c} \cdot b^{c+a} \cdot c^{a+b}$$
Taking $\log$ on both sides, we get $$\log{l}=(b+c) \cdot \log{a}+(c+a) \cdot \log{b}+(a+b) \cdot \log{c}\tag{5}$$
Now, using $(2),(3)$ and $(4)$ in $(5)$, we get
$$\log{l}=k(b^2-c^2)+k(c^2-a^2)+k(a^2-b^2)=0\tag{6}$$
From $(6)$, we get $\boxed{l=e^0=1}$
Roby5
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@Did Technically different as we find the value of $\log{l}$, from which we find the value of $l$. So this offers a different insight. Moreover, many answers on this site are copied from others, which don't offer any insight on how to approach the problem. Still they are voted upon and praised by some members of this community. Why are they not criticised? – Roby5 Jul 09 '16 at 19:02
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@Did To validate my claim see Muralidharan's answer posted here and many similar examples. Thus, I will not delete this answer, if that one is not deleted. – Roby5 Jul 15 '16 at 11:41